All categories

3 years ago

A cross-country skier can complete a 7.5 km race in 45 min. What is the skier’s average speed?

Physics

1 answer:

kap26 [50]3 years ago

3 0

Answer:

Speed = 10 km/hr

Explanation:

Speed = Distance / Time

Since we have to convert it into minutes to hours ( because it is km/hr )

45 minutes = 0.75 hours

Therefore,

Speed = 7.5 / 0.75

= 10 km/hr

Hope it helps!

You might be interested in

If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?

HACTEHA [7]

$\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time, t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0 +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}$

5 0

2 years ago

The pivot point of a lever is called the fulcrum. please select the best answer from the choices provided t f user: the thread a

kkurt [141]

It is TRUE. The pivot point of a lever is called the fulcrum .

4 0

3 years ago

Read 2 more answers

How many layers does the world have?

Lemur [1.5K]

The world has a total of 7 layers

6 0

3 years ago

Read 2 more answers

A hockey puck of mass m1=165 g slides from left to right with an initial velocity of 15.5 m/s. It collides head on with a second

Fynjy0 [20]

Answer:

it may be -6 m/s

Explanation:

3 0

3 years ago

Consider a 30-cm-diameter hemispherical enclosure. The dome is maintained at 600 K, and heat is supplied from the dome at a rate

SIZIF [17.4K]

Answer:

$\epsilon_2=0.098$

Explanation:

Diameter $d=30cm=0.3m$

Temperature $T=600k$

Rate of supply $r=65W$

Emissivity of base surface $\in_b =0.55$

Temperature at base $T_b=400k$

Generally the equation for Area of base surface is mathematically given by

$A_b=\frac{\pi}{4}d^2$

$A_b=\frac{\pi}{4}0.3^2$

$A_b=0.0707m^2$

Generally the equation for Area of Hemispherical dome is mathematically given by

$A_h=\frac{\pi}{2}d^2$

$A_h=\frac{\pi}{2}0.3^2$

$A_h=0.1414m^2$

Since base is a flat surface

$F_{11}+F_{12}=1$

$F_{11}=0$

Therefore

$F_{12}=1$

$A_b=0.0707m^2$

Generally the equation for Net rate of radiation heat transfer between two surfaces is mathematically given by

$Q_{21}=-Q_{12}$

$Q_{21}=\frac{\sigma(T_1^4-T_2^4)}{\frac{1-\epsilon}{A_b\epsilon_1} +\frac{1}{A_bF_{12}} +\frac{1-\epsilon_2}{A_h*\epsilon_2} }$

Where

$\sigma=5.67*10^{-8}$

Therefore

$65=\frac{(5.67*10^{-8}(400^4-600^4))}{\frac{1-0.55}{0.0707*0.55}+\frac{1}{0.0707}+\frac{1-\epsilon_2}{0.1414*\epsilon_2}}$

$\epsilon_2=0.098$

$\epsilon_2 \approx 0.1$

Therefore the emissivity of the dome is

$\epsilon_2=0.098$

3 0

3 years ago